Question: $g(x) = -6x^{2}+3x-2(f(x))$ $f(n) = -6n$ $h(x) = 2x^{2}+3x+7+2(f(x))$ $ g(h(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = 2(1^{2})+(3)(1)+7+2(f(1))$ To solve for the value of $h$ , we need to solve for the value of $f(1)$ $f(1) = (-6)(1)$ $f(1) = -6$ That means $h(1) = 2(1^{2})+(3)(1)+7+(2)(-6)$ $h(1) = 0$ Now we know that $h(1) = 0$ . Let's solve for $g(h(1))$ , which is $g(0)$ $g(0) = -6(0^{2})+(3)(0)-2(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = (-6)(0)$ $f(0) = 0$ That means $g(0) = -6(0^{2})+(3)(0)+(-2)(0)$ $g(0) = 0$